WHAT Examples (25-Jun-15)

WHAT,
Wolfberg's
Helpful
Anagramming
Tool, is really a toolbox of powerful tools,
with many facilities. You can make effective use of
WHAT right away by using just a few of its features,
and the document entitled
"WHAT's First" introduces these
features.
Many users will find these basic features are all they need to employ to
benefit by its use, but
WHAT provides a wealth of other facilities you ought
to explore when you have a chance. One way you can do this is by
studying the examples in this document.

These examples were updated in June, 2015 to be based on the OWL3 lexicon.

1. User Guide Examples

1.1 Is there at least one 7-letter word with AEILNOS?

If you ask WHAT to perform this command:

aeilnos~

it answers with:

There is at least one word.

and perhaps you are thinking it is
SEALION. If so, you would
be wrong, since there is a space in the two-word spelling of that
seallike animal. In case you were wondering,
SEALLIKE is acceptable in the
OWL3 lexicon.
You can ask to see the answer by requesting to WHAT that
it present the words on the slate with the one character command of an
apostrophe.

1.2 What are the highest-scoring opening plays with AEFKNOT?

You can ask WHAT to report the 10-highest opening
horizontal opening plays with one query with a few subcommands. Here is
the query and its resulting presentation:

If you type the above command, before you press the Enter
key, look at the Scoring tab in the upper right corner
of the WHAT window and both the
Query and
Presentation tabs at the left. You will see the meaning of
the command by the yellow-colored areas. Recall that you need not have
had to construct the command from your knowledge; the WHAT
GUI can be used to construct the command for you.

These are the individual subcommands in the above command:

/|2

-

filter based on word lengths - they are at least 2

/@*~

-

score for a play which include the center square

/<$

-

sort the presentation by decreasing scores

$

-

include scores in the slate presentation

10'

-

present the first 10 words only

1.3 Seeking a Triple-Triple Bingo

With the rack
ACILMT?,
what are the possible triple-triple bingos through an
R
in the 4th character position? There is indeed one such word, which
you can see in the following query and resulting presentation. The query
is a pattern match kind. It begins with a pattern followed by a comma,
which is the separator character you use to then present your rack. The
rack contents are used in filling in the missing letters in the pattern.
Here is the pattern match query with a rack:

2.2 Word is Contained In

You might want to know those longer words, such as 7's or longer which
contain a certain string inside (not starting or ending the longer word.
Let's say the word is LWEI
and we are looking for 6's or longer:

*?LWEI?*

The answer is:

EDELWEISS EDELWEISSES

Try the same kind of query for the base word of
POLICE. If you have not
yet seen this before, you may be surprised at the result.

2.3 Four I's

Someone at the Lexington Scrabble Club picked up a rack with four
I's,
and we considered what are the possible bingos.
WHAT can do this in several ways, the simplest of which is:

4I/S

which means what are the shortest steals where rearrangement is not
required. If rearrangement is required (using the command
4I/~S),
there are no answers, since they all have four
I's, with other letters
inserted.

A second way to do this is to try various numbers of blanks until there is
an answer. You might try
4I4?'s and see if anything
comes out, and then
press the F1 function key
to watch more blanks get added, one for each press. If you want
to see all words with four blanks, you can do it this way:

4I!*/A

The exclamation point forces the
four I's
to be in the answer. The asterisk
means any number of blanks, and since the asterisk implies pattern
matching, you then include
/A
to make the query kind anagramming.
It turns out there are more than 1000 such words in
WHAT's OWL3 lexicon.

2.4 SATINE plus a Blank

SATINE
plus a blank produces many words; and how many is that? With the query:

SATINE?#

The answer is:

Number of words = 73

But how many different letters can the blank be? The easiest way you
can ask this is with this query:

SATINE?#=

and WHAT reports:

The one blank can be any of 24 different letters.

Another way would be to bring back the previous command by pressing the
up-arrow key and then suffix the subcommand
/Q;W
to yield words in canonical order. This query results in:

Number of canonically-ordered words = 24

Although you can type the
=
command to see what letters the blank can
be and by observation find the 2 missing letters, here is how you can
explicit list those letters:

/1U=(A) // define set 1 to be one each of all letters
/2U=(=) // define set 2 to be those letters the blank matched
/3U=(1)-(2) // define set 3 as the difference between these sets
(3) // show set 3

You will see that the letters are
Q and
Y.

2.5 All about ACDEINT

Steve Hartsman posted a message to CGP about the rack
ACDEINT.
Steve listed various data relating to that rack.
The following WHAT output
shows how you get the same info using the program:

WHAT
is command language oriented, and there is a graphical interface
which can optionally be used to construct commands. In the above excerpt from
the WHAT workspace,
you see the prompt by the program on several lines, which is
"WHAT?: ",
and the remainder of the command line is the command you typed.

In the first command, as a result of your
typing the 7 letters and pressing
the Enter key,
WHAT came back with the response of
" (nothing)"
colored red, which means there are no 7-letter words with those 7 letters.

To form the second command, you could have pressed the
up-arrow key to get
back the previous command and then suffix these four characters:

?

-

meaning a blank

<

-

meaning "show front hooks (in lowercase)"

>

-

meaning "show back hooks (in lowercase)"

:

-

meaning "present the output using colons"

As a result of your then pressing the
Enter key,
WHAT presented the multi-line answer,
where each letter matched by the blank is shown in a distinctive color
(green); for example, on the final line the
U
in INCUDATE is green.

In the third command, you typed the equal sign followed by
pressing the Enter key,
which caused
WHAT to report on what the blank in the previous query
can be.

To form the fourth command, you pressed the
up-arrow key thrice to bring
back the original command, and then you
suffixed a U
and quotation mark. When you pressed the
Enter key,
WHAT presented the definition for
INCUDATE. You can then
use the mouse to double-click
"incudal"
to see its definition in the Definitions area above.

2.6 5-Letter Q Words with Anagrams

The following query causes WHAT to present words with
a Q which have anagrams:

Q4?/&>1

2.5 AEIOU 8-Letter Words

To get the 8-letter words with words which have the vowels
A,
E,
I,
O,
U
and 3 consonants, give this query to WHAT:

2.6 7's and 8's That Extend with TY

From John Van Pelt's Verbalobe web site,
there was a list of those 7- and 8-letter
words which can be extended with TY
There are seven 7-letter words and five 8-letter words
which have this characteristic. Here is how to ask WHAT
to present the 7's:

7.\ // first get all the 7's to the slate
[]ty // when TY is suffixed, what are words?

The 8's can be done in the same way.

2.7 What words end in either
ED or
T

From John Van Pelt's Verbalobe web site,
there are 64 words ending in either
ED or
T:

*ed#

says there are 14652 words, and

*t#

says there are 9448 words.
So begin with the smaller second group, and then
ask WHAT to perform:

[1-2|]ed

which yields 366 words.

2.8 Musical Words

To find out what words with at least 5 letters which can be played on a musical
instrument, namely made from letters in the range
A -
G, use this query:

10(A-G)/|5

2.9 Collins 2's

Collins is the lexicon which includes the words from North America
and Great Britain. This lexicon is used in most countries
throughout the world and for world competitions,
but it has not yet been accepted for all competitive games in North America.

Perhaps you want to look at 2-letter words from the Collins 2012 Lexicon.
You probably would like to be shown which of these are from the
British list only, such as with a suffix of a number sign. You must use
the WHAT graphical interface to alter the lexicons - there are no
commands to change the lexicons for you to present in the workspace.

Select the Lexicons
tab in the upper right corner of the WHAT window.
Click on the Change Lexicon(s)
button. A dialog appears a the lower
right corner of the
WHAT window. Select "Collins Words 2012 ..." as the primary lexicon
and "OWL3 ..." as the secondary lexicon at the bottom of the dialog.
In the upper part of the dialog, select the middle option which says to
identify words NOT also in the secondary lexicon with some suffix
character. Then go the right area of the dialog and select a number
sign as the suffix character. Click the OK button and you are now
going to see answers to queries coming from Collins, and words not
in OWL3 will be suffixed with a number sign. For example, take a look
at the 2-letter words with a command such as two periods; you are
then shown:

You can see those 19 2-letter words acceptable only from Collins 2012.
If you want to make a list of just
these words, save the slate to a wordlist. Change the lexicon settings
to produce answers from OWL3, get the OWL3 twos and remove these from
the wordlist you saved.

2.10 Random Rack Subwords

As suggested by Joe Edley, let's say you have produced a random rack,
and it has been input as a query. What can you now do to find out its
subwords?

For the random rack of
UMDNEEO, you are shown the
answer of EUDEMON.
Now, press the up-arrow key and then hold down the
Ctrl key and press the digit
3.
You will be shown all those words of at least length
using the seven letters in
UMDNEEO.

When you hold the Ctrl key and press
3,
these command characters are suffixed to the query:
" /|3".

2.11 ...EABLE Words

Wendy Wolfberg asked how to spell "upgradable",
and I first typed that
in to find it is a word. Then I pressed the up-arrow key,
typed an E,
and pressed the Enter key
to see if it could be UPGRADEABLE,
and indeed it is also a word. I did the same for
GRADABLE and found the
E there is
not allowed. So, I wonder what other words of this form can have the
optional E. Here is
some WHAT input and output:

*eable# // get all EABLE words onto the slate; see how many

The answer is 139 words.

[1-6|]able // for each word on the slate get all but its final
// 5 letters, add ABLE, and which of these are words?

3. Simple Questions with Tougher Answers

3.1 What words have both a
J and a
Q?

What words have both a
J and a
Q?
The easy command to ask for this is:

j!q!*/a

The suffix exclamation points indicate that
J and
Q
are required to be in the answer. The asterisk represents
any number of blanks, but since it also implies the query is a pattern
match kind, you must also change it to be an anagram kind of query, and that
is what the /a does.

This is a indeed a pattern match query for all words, and the rack
includes the two required letters of
J and
Q
plus any number of other letters.

3.2 7-Letter X Words with Anagrams

Issue this query:

x6?/&>1 // get the 46 7-letter words with anagrams

Now ask WHAT to present these in pairs. One way you can
do this is to ask it to present output in rows with two columns and
sort by canonical order:

/ORC9W /OC2C />; // produce the desired answer

3.3 Z.....S Words Requiring that S

What 7-letter words starting with a
Z
and ending in
S
cannot drop the
S
to form a 6-letter word?

John Chew listed these answers for the TWL98 Lexicon:

zealous zilches zincous zipless zloties zygosis zymoses zymosis

and ZAIDIES
is new in OWL3.

To do this in WHAT:

z5.s\ // get the 7's with start with Z and end with S
/1w // save to wordlist 1
[1-6]\ // get the 6's which are acceptable when dropping the S
[]S\ // put the S back on these 6's
/y0=1-0 // produce the answer by taking the difference
' // present the answer of 9 words

3.4 No S Back Hook Words

Michael Eldeiry announced he was seeking a list of all words of length
2 through 8 which do not take an
S
as a back hook. There are several
ways to do this in WHAT,
and here are two of them. Beware that the answer
is rather large, so you may want to do this work in pieces. For example,
let us begin with just the 3-letter words.

Get words onto the slate for which you want to ask this question. If
you first just want a list of the threes, issue the command:

3.\

Then make the query using these words as input such that the retained words
are those set of back hooks is a subset of the set with all letters but
S:

3.\ // Start with the candidate words, such as the threes:
/1w // save these to wordlist 1
[]s\ // create the fours which are S extensions of these threes
[1-1|]\ // remove the S, so these are the threes with take S
/y0=1-0 // take the set difference (all minus those which take S)

If you want to make the full list Michael asked for, start with
the following command to have all words of length 2 through 8:

*/|2-8\

This is 88099 words. Once again the second query to yield the final answer is:

These above two commands yield those 6-letter words which consist of a
repeated 3-letter word. But notice a word like
ATLATL is missing,
since ATL
is not a 3-letter word. There are at least two methods to
get such words:

You can start with this command:

3./qs\ // get all 17576 3-letter strings onto the slate
2[] // get words which consist of 2 of the same string

There are 33 such words, 15 of which are the ones formed with repeated 3-letter words.

You can use this somewhat different approach:

6.\ // get all 6-letter words onto the slate
[1-3]/QS\ // for each 6-letter word, yield a 3-letter string (not
// necessarily a word) from the first 3 letters
2[] // for each of those 3-letter strings, yield words
// which consist of 2 of these strings

Assume the answer of the above 33 words, including
ATLATL, is on the slate.
Now, let's find which of these have a consonant as the second letter.
Here are two different methods of doing this:

First, the more convoluted method, takes three steps:

/1w // save the slate to wordlist 1
.(C)4. // get all 6-letter words with a consonant as letter 2
/y0=0&1 // yield the intersection of the 2 lists

A simpler way to do this is to use the slate as the source of words
and do a pattern match. Again, assume the 29 words are on the slate:

/0I .(C)4. // get the 6's with a consonant as letter 2

This is the answer:

ATLATL CHICHI GRIGRI GRUGRU TSETSE TSKTSK TZETZE

3.7 6's Made from Two 3's

What 6-letter words consist of two consecutive 3-letter words, such as
STYRAX?

There are many ways to approach this, and this is just one of them:

3.\ // get all the 3's
[]3./1w\ // for each, suffix 3 letters to form a word; save it
3.\ // get all the 3's again
3.[]\ // for each, prefix 3 letters to form a word
/y0=0&1 // intersect the two lists
28' // present just the first 28 of these

3.8 8's Made from Two 4's

What 8-letter words can be formed by concatenating two 4-letter words?

WHAT
does not have a way to iterate over two lists, but it can be used to
produce the answer. Here are two completely different ways:

Method 1:

8.\ // get all 8's
[1-4]/#=1 \ // yield those 8's which begin with a 4
[5-8]/#=1 \ // from previous, yield those 8's ending with a 4
/0I z7. // use slate as source; get those starting with Z

Method 2:

4.\/4w // put all the fours into wordlist 4
Z7.\ // get all the 8's starting with Z
/4I[]/2p // using the 4's as the word source, do 2-word pattern matching
// with the words on the slate (the 8's)
[] // combine the pairs of 4's into 8's

3.9 Find all the 4's in
OATEN

You are presented with a quiz to find all the 4-letter words in
OATEN.
You think you have the answer, and you want to check it to see if you
found all the words. Make this probe to find out how many words are in
the answer:

OATEN/|=4#

In the above command, the subcommand
following the letters is a filter that the word length must be equal to
four. But if you are unsure of the query you made, and you want to
reassure yourself the words just counted are indeed of length four,
show the slate with the words hidden and with word lengths using
this command:

]|

You could have had the
WHAT GUI make up this command by going to the
Presentation
tab on the left side of the WHAT window, and clicking the
radio button labeled "Hidden, using [] brackets" and the option button
labeled "Lengths". The presented result is:

[].4c []_4c []_4c []_4c []_4c []_4c []_4c []_4c

which should be reassuring that there are indeed 8 4-letter words on
the slate. If you want to check whether a word you have in mind is
indeed a word, type a command such as:

eton/-?

are you will be told whether
ETON
is an acceptable word. You could at
this point, ask whether a word is on the slate, such as:

nota/?

You will be told just whether
NOTA
is on the slate. If you instead ask:

nota/0?

when the word is on the slate, you are also told at what that item number
the word is; to make that datum have no meaning you could first shuffle
the order of the slate using the command:

/0S

where that second character is the digit zero.
But when you then ask for whether a word is on the slate, you should
turn off alphabetical sorting, so you could issue the command:

nota/0?/='

The final subcommand in the above command could be created for you if
you go to the Sorts area of the
Order By box at the bottom of the
Presentation
tab and select "not by alphabetical order". In other
words, you do not have to remember the command syntax - the command
can be created for you.

4. Questions for Word Study

4.1 High Fives

In "The Complete Wordbook for Game Players",
author Mike Baron suggests
there are so many 5-letter words that one should approach these in
groups. After studying those with high-scoring letters, it then makes
sense to study those with a first or fifth letter of score 4 or 5.
Mike calls these "The High Fives". This list of high fives,
which can be made by WHAT,
is rather large, and you may want to break it up into
smaller groups, such as those high fives which begin with
A.
Here is the command to get these words:

A3.($4-5)

There are 48 such words. If you would like to present them with related
info, such as anagrams, WHAT
can do this for you, and here is how this
looks for those high fives starting with the letter
O:

4.2 GAYWINGS

Joey Mallick mentioned that
GAYWINGS
a perennial herb) must have
the final
S.
We wondered whether there are other wuch words ending
in "WINGS".
This is an example of when to use the "unhooks" feature
of WHAT. Ask
WHAT to show unhooks, and you can see that indeed that
is the only such word. With this query:

*wings`

where that final character is the back quote or accent grave, the
result is:

All of these words other than
GAYWINGS have a suffix of
"-",
which in an indicator you can unhook the final letter.

If the list of
...WINGS
words were longer, you might prefer to answer
this kind of question in a different manner. Here is another way to
accomplish this:

*wings\/1w // collect all the words and copy them to wordlist 1.
[1-2|]\ // for each of these words, keep those which can drop S
[]s\ // for each of these, put the S back on
/y0=1-0 // take the set difference, and output to the slate

The output from the last command indicates the slate has one item. To
see it, issue the one-character command of an apostrophe.

4.3 Highest-Scoring Triple-Triple 8's

Here is a way to find the highest-scoring triple-triples of length 8:

/@1A.! // scoring will be based on starting at square 1A
8./$350/>'/<$$/OC18X // filter based on scores at least 350,
// sort by decreasing scores & alphabetical

4.4 Highest-Scoring 21-Letter Words

Find the top-scoring 21-letter words when the horizontal placement
of the word starts at position 1A on a board for the
21 x 21 board game:

/21B // set the board size to 21 x 21
/@1A.! // score horizontal words which start at the upper left square
21./$>8400/>'/<$ $ /o1 // find all 21-letter words which score more than 8400, sort
// by decreasing scores and show with scores

4.5 The Blank Must Be an X

Knowing that
UXORIAL
is the only 7-letter word which can be made from
U?ORIAL,
what other 7's have this characteristic? There are 76 such 7's.
Do any of them have anagrams? The answer is "no".

x6?/Q;W // get the 535 7-letter canonically-ordered words with X
[]-X+?/:=(X)/Q;W# // get the 78 canoncially-ordered words that take only an X
[]/A // show these words
/<% // show these sorted by decreasing probabilities,
// seeing UXORIAL is first; perhaps some of these are
// unexpected, such as PTYAEI+X
[]/&>1 // look for those with more than 1 in the anagram set,
// and there are none - perhaps not surprising

4.6 5's Not Contained in 8's

Joev Dubach asked for a list of the 5's which are not contained in any 8's.
Here is how you use WHAT to produce the answer:

5.\/1w // get all the 5's and put then into wordlist 1
8.\ // get all the 8's.
[]/a/|=5\/2w // for each 8 find all 5's and put them into wordlist 2
// (this takes 2-3 minutes to complete)
/y0=1-2 // take the difference, and this is the answer

4.7 Convert to Alphagrams

Steve Hartsman asked for a way to take a given list of words and
convert each of these into alphagrams. What was unclear is if the
order of the list should be retained, and if not should duplicates
be removed.

If the order of the list is to be retained and duplicates kept, then
import the given list to the slate, clear the workspace via the GUI or by
using the /CW command,
and then present the slate with canonically-ordered words and requesting
one word per line with no sorting (alphabetically), namely:

/o1/=';

Then output the workspace to a file; you will have to remove the first
two lines (the command and a blank line) and the final line (a blank line)
outside of WHAT. This is then the result.

On the other hand, if the order of the list is not to be retained and you
are removing duplicates, then after importing the input list to the slate,
execute the command:

[]/q;w

Then you can export the slate to a result file; there are no extra
lines which must be removed with this method.

5. Curiosity Questions with Simple Answers

5.1 7's Which Match Their Alphagrams

An alternate way to express the notion of words which match their
alphagrams is: "What words are spelled in alphabetical order?".
This assumes the current canonical ordering is the default of alphagrams.
Here are the commands to do this:

7./Q;W\ // get canonically-ordered 7-letter words; don't show them
[] // for each, yield only words (pattern matching is implied)

The answer is merely BEEFILY and
BILLOWY.

5.2 7's Which Match Their Reversed Alphagrams

In a message posted to CGP, Steve Hartsman asked what alphagrams
of words of at least 7 letters spell that word reversed. His example
was TRIGGED.
This is similar to the previous example. All you need
do is define the canonical order to be the alphabet in reverse.

/;"ZYXWVUTSRQPONMLKJIHGFEDCBA" // define this canonical order
7./Q;W\ // get the canonically-ordered 7-letter words, but don't show them
[] // for each one, yield only words (pattern matching is implied)

Do this for 8's, and TROLLIED
is the only answer, and that is the
longest. There are 54 such words of length 6.

5.3 ZINGARA-like Words

We have noticed the word
ZINGARA
can end in any of
A,
E,
I, or
O.
We wonder what
other 7-letter words have this characteristic, where the only
final letters are those four vowels. Here are the commands to
find out:

6.a\ // get all 7-letter words ending in A
[1-6]./:=(AEIO) // get those 7-letter words ending in A whose last
// letter can be A,E,I,O only

This yields only
STRETTA and
ZINGARA, which answers the question.
Even if we change the = to >= the same answer is produced, which is
not surprising.

5.4 11's With 3-Anagram Groups

Aaron Bader thought it was interesting to see the 11-words which are in
3-anagram groups. Here is a way to get the list of alphagrams for these
words. Then you can use these as the basis for a quiz. Aaron suggests
the most interesting group is ACEHOOPRRST.

One more technique to get the same list is to seek wotds with a
probability of zero, since probabilities are calculated based on
letters without any consideration to blanks. This command does the trick:

4./%=0

6. Curiosity Questions with Tougher Answers

6.1 Words With Five Vowels in Alphabetical Order

What words contain
A,
E,
I,
O. and
U, where these
letters show up in alphabetical order within the word?

First, find which ones begin with
AB
and may have other vowels too:

a*e*i*o*u*

These 5 words are shown:

ABSTEMIOUS ABSTEMIOUSNESS ABSTENTIOUS
ABSTEMIOUSLY ABSTEMIOUSNESSES

Then, restrict the question to
words which start with
A,
where those five vowels are the only vowels:

6.2 Getting Hints for a Bingo Rack

You are provided with a rack in which there is a bingo, and you want
some hints. You can ask for specific letters of specific words, but
perhaps you want to just know whether a certain letter is involved.
For example, you may want to know whether the final letter of any of
the words in the answer is a
C,
or you may want to know this for a
specific word, such as the first one. You can do make such a query, but
it may modify the slate, so it would be advisable to copy the slate
to a wordlist before making the query, and then you can restore it
when you want to. The following query asks whether the 7th letter
of any of the 7-letter words on the slate is a
C:

[1-6]C~

If you want to know the same thing for the fourth letter:

[1-3]C[5-7]~

Having said all this, there is still the possibility that when alternate
source is supported more than for subwords, this will make such a query
easier. Let us assume a pattern match query can use source words from a
given wordlist. Then:

/1W // save the slate to wordlist 1
/1I! // use wordlist 1 as the source (long-term)
6.C~ // are there any 7-letter words in wordlist 1 which end in C?

6.3 C-G-P Words

John Morse posted to the CGP Email List some letter sets with one blank such
that the blank can be exactly
C or
G or
P.
WHAT can be use to find such answers:

6?c# // get all 7's with a C
[]-c+g/#>0# // from those words, keep those which produce a word
// when a G replaces the C
[]-c+p/#>0# // from those words, keep those which produce a word
// when a P replaces the C - there are 569 words
[]-c+?/:=(CGP) // from those words, keep those which produce words
// such that the blank is exactly C,G,P

Two words come out: VINCULA and
WACKILY.
Then, if you want to see these in canonical order:

[]/q;w // yield these answers in canonical order
[]-c/qs // yield strings from these answers less one C

6.4 8's With No 7-Letter Subwords

Avi Moss asked for the 8's with no 7-letter subword, i.e. a 7-letter word
prefixed or suffixed with a hook letter which leads to an 8-letter
word is not included. Here are WHAT to find these:

8.\/1w // make a list of all 8's and put into wordlist 1.
7.\ // make a list of all 7's
.[]\/2w // make a list of all 8's which are formed from a 7 with
// a front hook, and put into wordlist 2.
7.\ // get the 7's again
[].\/3w // make a list of all 8's which are formed from a 7 with
// a back hook, and put into wordlist 3.
/y4=2|3 // combine wordlists 2 and 3 into wordlist 4
/y0=1-4 // yield the answer onto the slate

The answer to this is rather large.

6.5 Minimal Longest Subwords of 8's

Seth Lipkin noticed from a word-of-the-day listing that
PANMIXIA's
longest subword (when anagramming) is of length 5. He asks are there
any 8's with smaller maximum length subwords? Here is how this can
be answered in WHAT:

In the presentation of the 63 strings, you are not shown which
are new words in OWL3. To see this, issue the command consisting of
an an open and closed square brace. You will see two of these words are new.

Then, Seth followed up by noticing in the above list that all the words
have repeated letters, and there is a small number of these with only
a single repeat, such as
DUMMKOPF.
We could first run a check to be
sure we did not miss something - which words have no repeated letters
(if any)? Here is one way to answer this in WHAT:

7.2 Longest Words With No Repeated Letters

This question arises on a regular basis: What are the longest words with
no repeated letters? This is very easy to answer using
WHAT, since there is a
set which represents a first-time letter of a word. The only problem is to
formulate a query in which you cover multiple word lengths. Let's say you
guess the answer is at least 10. You can ask for words of length to the maximum
length of 26 with this query:

If you are then curious what is the last word in the list, you can type
the command:

/>| /1 // sorting by increasing length, what is the first word

You will be shown it is
ABDUCTIONS; so this is the alphabetically
first 10-letter word with no repeated letters.

7.3 Longest Words With No 1-Point Consonants

Steve Root asked what is the longest word with no 1-point consonants.
Of course, that would be these consonants only:
BCDFGHJKMPQVWXZ.

Two methods are included here:

This first method does not use sets.
Make a query with these letters replicated 5 times and augment this
with 10 Y's
and 10 vowels, and ask for words longer than length 6. You will be told there
are 1828 of them, so change the 6 to an 8, and find out there are 443 of these.
Finally, change the number to 11, and you will find that
DEHUMIDIFIED is the
unique 12.

This second method uses sets:

/1U=($2)|(V) // define set 1
(1) // look at set 1 to be sure
0(1)/<||12' // query for any number of set 1 letters, and
// sort by decreasing length, show length, and
// the first 12 words

7.4 Making 9's With Only
DW and
UW

When Steve Root was doing a postmortem on a Clabbers game he played,
he noticed with whatever word tools he had at his disposal that his
rack made 9's with
DW and
UW only.
He wondered how many other 7-letter
racks have this characteristic. WHAT came to the rescue:

You might be tempted to combine the final two steps above, using the
following query:

*,[](+)(R-1)(R-1) // get words of the form ABBCCC anagrammed

but, at least for now,
WHAT does not support query-based sets in racks.

7.6 6+2 Unistem Words

Seth Lipkin mentioned he likes the 6+2 Unistem list of John Chew. A
"unistem" is a set of letters that, together with a certain number of
blanks, anagrams to form only one word. For example,
HOAGIE is a
two-blank unistem using the OWL3 lexicon, because the only 8-letter word in
AGEHIO?? is
ESOPHAGI.
Since HOAGIE happens to be a word itself,
rather than a random collection of letters, it is a 6-letter two-blank
unistem word, or a 6+2 unistem word for short.
John presents these in reverse probability order on his web site, and
here is how you can produce this list using WHAT.

6.\ // get all the 6-letter words onto the slate
[]+??/#=1\ // replace the slate with only those 6's which lead
// to one 8-letter word
/<% 24' // present the first 24 words of the answer

John Chew provides the answers as 6's and in decreasing probability;
WHAT's first 24 such answers are (read down columns first):

John's list is in a slightly different order, since his probability computation
includes taking blanks into account, but WHAT's computation
does not.

8. Esoteric Questions with Tougher Answers

8.1 7's With No Anahooks

Matthew Hodge asked for a list of the highest probability 7-letter
words which have no anahooks, such as
Since NOTATES.
This means that Since NOTATES
plus a blank yields no eights.

Begin by making such a list and then decide how many make sense:

7. /Q;W # // get all racks for 7-letter words

After about 4 seconds of working, WHAT announces the
number of these is 21063. Now prune this down to those which have no anahooks:

[]? /#=0 #

After working about 8 seconds, WHAT announces
the number of these is 4136.

Set the presentation of the slate for the long-term so that answers
are sorted by decreasing probability and show nothing:

/<%! \

It is worth noting that you do not have to make the commands from your
memory and knowledge of WHAT.
You can use the GUI to formulate these
commands. The first two of the above three commands are constructed
using choices you make from the Query tab,
except for the terminating
number sign, which you can get from the Presentation
tab, but you just
might recall that particular subcommand without using the GUI. The final
command is constructed using the bottom part of the
Presentation tab.

Now see what the 190th rack is, and show the words (technically, the
anagrams of the answer):

/200 )

The presented answer is:

AEINTXY (ANXIETY)

It might make sense to save this list now to a wordlist. For example,

/1W

saves the 4136 racks to wordlist number 1. Now if you want to work with
only those racks which are of the probability of
ANXIETY or greater, one
way to do this is start over and include the filtering subcommand of:

/%{ANXIETY}

in your first query. Alternatively, you can direct
WHAT to use a word list
as the basis of words for a query which includes filtering. Let's detail
both of these sub-scenarios:

Don't start over, but use the
current list as the basis of a query and filter:

/0I 7. /Q;W /%{ANXIETY} #

The second character in the above command is the digit zero, and that
value of zero indicates the slate is the word source for this command.
WHAT reports there are 202 words.

The second way of doing this is to start over with these commands:

7. /Q;W /%{ANXIETY} #

The answer is 4889.

[]? /#=0 #

The answer is 191.

Either of the two sub-scenarios get you to this point.

Now, look at the first 50 of these, since that is about how many can be
shown in the workspace at once (My workspace is 30 rows of 89 columns.).
Show both the number of ways to make the rack (the basis of the
probability) and the word(s):

You could export the wordlist of these 191 racks to a
file and use it as the basis for flashcarding.

8.2 8's Which are not Extensions of 7's

Bob Gillis would like to get
the top 1000 8-letter words which do not have a 7-letter word as a subword
by removing the first or final letter. An example is
SALMONID. Perhaps a
more interesting set of words are those 8's which have no
included 7-letter words when anagramming, but here is how to answer Bob:

8.\ // get all the 8's
[]/a/|=7/#=0 // keep only those 8's with the sought characteristic

From this, one could present a quiz: what group of words of length 8
has the most number of anagrams and there are no 7's in the letters?

8.3 Symmetrically-Distributed Letters

There is a web site with considerable info about words. One category
there was for words with symmetrically-distributed letters, such as
IZAR and
WIZARD. Notice
I and
R are symmetric in that
I is the ninth letter of
the alphabet, and
R is the ninth last letters
of the alphabet.

WHAT can help seek these words. For example, find
the 6-letter words with this characteristic:

8.4 Longest Words with Imbedded 2's, etc.

From John Van Pelt's Verbalobe web site,
a question was posed: "What are the longest words with all
imbedded 2's?"
That was then extended to ask the same question for other lengths up to 7,
and these are the answers:

2's

from

11's

-

DENOMINATOR + 9 others

3's

from

10's

-

CALABASHES only

4's

from

8's

-

PARAKEET + 2 others

5's

from

9's

-

TRAVERSES and 3 others

6's

from

8's

-

BLENDERS + 12 others

7's

from

9's

-

CAROUSERS + 2 others

To find such words using
WHAT, you need to make
individual probes for specific pairs of lengths.
For example, to find the longest 9-letter words with the most
imbedded 5-letter words, use these commands:

These three words result:
HURTFULLY,
SPRIGHTLY, and
TRIGLYPHS.
By inspection you can see that the sought answer consists of the
2nd and 3rd words.
RUTHFULLY
is not in the list,
since it has
the 2-letter word UT
imbedded within.

In order to check for the number of anagrams in the answer and not
in the lexicon, suffix the final command above
with /1I to use
wordlist 1 as the word source instead of the lexicon. In this case,
only the two words are in the answer.

8.6 The Most Imbedded Words

Someone mentioned that
THEREIN
contains many shorter words spelled in order.
WHAT can be used to check whether this word has the
most imbedded words for all 7-letter words:

therein/i# // find out how many words are imbedded in THEREIN

The answer is 13, which includes the word itself.

7.\ // get all the 7's onto the slate
[]/i/#13 // get those with more than 12 imbedded words
[]/i/#14 // get those with more than 13 imbedded words
[]/i/#15 // get those with more than 14 imbedded words
[]/i/#16 // get those with more than 15 imbedded words
[]/i/#17 // get those with more than 16 imbedded words
[]/i/#18 // get those with more than 17 imbedded words
[]/i/#19 // get those with more than 18 imbedded words

So now the answer of 11970 words is in wordlist 1. These, of course, are
not all truly compound words. Realize the answer has only 8-letter words.

Another way to go about this is to use the query kind of 2-word patterns,
but with this kind of query, there is no control on the length of the
subwords. Commands to do this are:

8. \
[] /2P #

One could take the answer of 19013 strings from
WHAT and remove those lines
which break the longer word other than what is desired. The lengths of
each of the two-word results can be used for sorting. For example:

/>|/8~W // sort by increasing word lengths and save to wordlist 8

At this point, the wordlist has the two-word results sorted by length. In
this wordlist, the first items have 6 letters as the size of the first
word, and the end of the list of 19013 items has 2 letters as the size of
the first word. The stated goal is to eliminate both of these groups from
the entire answer. We have to do is find where those groups start and stop.
A way to probe wordlist 8 is via the dialog which comes up for wordlist
item deletion - menu pick
Tools→Wordlists→Delete Items... to bring up
the dialog. You can examine items of wordlist 8 and home in on where
the items change length. Using this information, you can use the dialog
to form the appropriate item deletion command.

With these probes I found the border between the length-6 and length-5
first words. So now, those first 3157 words can be deleted from the
slate using the delete wordlist items command:

/=8:1,3157D

Then, the same item deletion dialog can be used to find the trailing
items to delete. At this point, these are items starting at 12906.
So issue a similar command:

/=8:12906,5000D

The count has to be at least how many items are to be deleted, and 5000
is higher than needed. So, at this point, 12905 items are now remaining
in wordlist 8. This number of items is higher than the number of 8-letter
words found by the first method described above, namely 11970. The
difference is that in wordlist 8, an 8-letter word may be showing up more
than once, since it can be split at more than one place. We can prove
this by running a modified wordlist 8 back through the anagrammer with a pattern
query. Export wordlist 8, use a text editor to remove spaces, and import
the result back into wordlist 8. Then, the following query should then yield 11970 words:

[]/8[# // using wordlist 8 as the source, pattern match each word

Yes, indeed it does - there are 11970 words.

8.8 8's With Four State Abbreviations

Using the 2-letter abbreviations for the 50 states plus
DC, what 8-letter
words consist of four state abbreviations concatenated? Here is a
way to answer this using WHAT:

9. Beyond the Power of WHAT

Steve Root came across a word whose 4 anahooks were among the letters of
the word; he subsequently forgot what that word was. He also wondered
what is the word with the most member anahooks.

Someone on CGP noticed word pairs such as
CENSURE -
MILORDS such that
the anahooks of the first word are those 7 letters in
MILORDS. Can WHAT
be used to find all such word pairs?

In fact, these are not necessarily
word pairs - the words in each side of the pair may have anagrams; for
example
CRASHED and
ECHARDS lead to
DIPTERA,
PARTIED, and
PIRATED.
This may be beyond the scope of WHAT's
capabilities without some other processing outside of
WHAT, but such
processing may be relatively simple, such as using a text editor. Here are
some thoughts.

Yield all 7-letter canonical racks which lead to words:

7./q;w#

and 21063 is how many there are. For each of these, add a blank and retain
those words for which there are at least 7 resulting words:

[]?/#7

and 2584 is how many there are.
Export this list to a file and add a
? to each line in the file.
Set the long-term kind presentation to
"What 1-2 blanks can be"
with:

=!

Clear the workspace. Execute commands from the file with the question
marks. Export the workspace. We are not there yet,
but we are on the way.